Must there be an invertible matrix $Q$ such that $Q^{-1}MQ=N$?

Question

Let $M$ be a complex square matrix such that it has the following:

Minimal polynomial: $\mu(\lambda)= (\lambda-3)^2(\lambda+2)^3(\lambda-5)$

Characteristic polynomial: $\chi(\lambda)=(\lambda-3)^4(\lambda+2)^5(\lambda-5)^2$

$\dim(E_3)=2$ and $\dim(E_{-2})=3$, where $E_\lambda$ is the eigenspace for $\lambda \in \mathbb{R}\,$.

Now, my question is this:
If $N$ is another complex square matrix with the same characteristic polynomial, minimal polynomial, and geometric multiplicities as stated above, must there be an invertible matrix $Q$ such that $Q^{-1}MQ=N$ ?

If this is true, I would like to understand why. And if not, a counter-example would be much appreciated.

Answer 1

All of the information that you have given can be interpreted as a description of the Jordan normal form of $M$. In particular:

Characteristic polynomial:

• $M$ has size $\deg(\chi) = 11$
• The sum of the sizes of the Jordan blocks associated with $\lambda = 3$ (i.e. the number of times $3$ appears on the diagonal) is $4$. This sum is $5$ for $\lambda = -2$ and $2$ for $\lambda = 5$.

Minimal polynomial:

• The largest block associated with $\lambda = 3$ has size $2$. Respectively, $3$ for $\lambda = -2$ and $1$ for $\lambda = 5$.

Algebraic multiplicity:

• There are $2$ Jordan blocks associated with $\lambda = 3$ and $3$ for $\lambda = -2$.

Let $J_k(\lambda)$ denote the Jordan block of size $k$ associated with $\lambda$, and let $\oplus$ denote a direct sum. Verify that with the information given above, the Jordan form of $M$ must be (up to a permutation of the blocks)

$$ J = [J_2(3) \oplus J_2(3)] \oplus [J_3(-2) \oplus J_1(-2) \oplus J_1(-2)] \oplus [J_1(5) \oplus J_1(5)]. $$

Two matrices are similar if and only if they have the same Jordan form. However, any matrix $N$ that has all of the listed characteristics must also have this Jordan form. It follows that for any such $N$ is similar to $M$, i.e. there is a matrix $Q$ for which $ Q^{-1}MQ = N$.

Answer 2

Warning: This answer is incorrect, as currently given.
Correction is underway.

If you are given both the characteristic and the minimal polynomial of a matrix, then you know its Jordan canonical form (which is usually not unique because permutations of Jordan blocks are still allowed).
A matrix $M$ and its Jordan normal form $J$ are similar, which means $Q^{-1}MQ=J$ for some invertible matrix $Q$.

Your question is answered in the positive, because $M$ and $N$, sharing characteristic and minimal polynom, then also share the Jordan normal form, and both $M,N$ are similar to it. Now recall that similarity is an equivalence relation.

If you strive for uniqueness, then look after the Frobenius normal form of a square matrix.